ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-2\\y\le\frac{16}{3}\end{matrix}\right.\)
\(2x^2-\left(3y-6\right)x+y^2-8y-20=0\)
\(\Delta=\left(3y-6\right)^2-8\left(y^2-8y-20\right)=y^2+28y+196=\left(y+14\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3y-6+y+14}{4}=y+2\\x=\frac{3y-6-y-14}{4}=\frac{y-10}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=x-2\\y=2x+10\end{matrix}\right.\)
- Với \(y=2x+10\ge-2.2+10=6>\frac{16}{3}\) ko phù hợp ĐKXĐ (loại)
- Với \(y=x-2\)
\(4\sqrt{x+2}+\sqrt{22-3x}=x^2+8\)
\(\Leftrightarrow x^2+8-4\sqrt{x+2}-\sqrt{22-3x}=0\)
\(\Leftrightarrow x^2-x-2+\frac{4}{3}\left(x+4-3\sqrt{x+2}\right)+\frac{1}{3}\left(14-x-3\sqrt{22-3x}\right)=0\)
\(\Leftrightarrow x^2-x-2+\frac{4}{3}\left(\frac{x^2-x-2}{x+4+3\sqrt{x+2}}\right)+\frac{1}{3}\left(\frac{x^2-x-2}{14-x+3\sqrt{22-3x}}\right)=0\)
\(\Leftrightarrow\left(x^2-x-2\right)\left(....\right)=0\) (ngoặc phía sau luôn dương)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\Rightarrow y=-3\\x=2\Rightarrow y=0\end{matrix}\right.\)
